Chapter 3 107 - CHAPTER 3 PROBLEMS PLUS 257 20(a f.r 33(216...

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Unformatted text preview: CHAPTER 3 PROBLEMS PLUS 257 20. (a) f(.r) : 33(216 — 2)(a: — 6) 2 m3 — 8w2 +123; :> f’(m) : 3m2 —16a: +12. The average of the first pair of zeros is (0 + 2)/2 : 1. At ac = 1. the slope of the tangent line is f’(1) = —1. so an equation of the tangent line has the form y = —1m + b. Since f(1) = 5. we have 5 : —1 + b :> b = 6 and the tangent has equation 0+6 2+6 y = —a: + 6. Similarly. at 1' = = 3. y : —9m + 18; at m = T = 4. y = —4a:. From the graph. we see that each tangent line drawn at the average of two zeros intersects the graph of f at the third zero. (b) A CAS gives f’(:c) : (m — b)(w ~ 0) + (:c ~ a)(;c — c) + (:1: — a)(av ~ b) or f'(ac) = 3:52 7 2(a + b + c)m —l— ab + ac + be. Using the Simplify command. we get 2 _ 2 f’<a:b> : _(a:1b) and f(a_2l—b> : _(a b) (a+b—2c).so an equation ofthe tangent line at y‘ (ahb)2<m (1:1)) (agb)2(alb 20) To find the m—intercept. let y : 0 and use the Solve command The result is ac : c. Using Derive. we can begin by authoring the expression (:5 — a)(a: — b)(a: ~ c). Now load the utility file Dif_apps. Next we author tangent (#1. m. (a —l— b) / 2)——this is the command to find an equation of the tangent line of the function in #1 whose independent variable is 1: at the ac-value (a + b) / 2. We then simplify that expression and obtain the equation y : #3. The form in expression #3 makes it easy to see that the sc—intercept is the third zero. namely c. In a similar fashion we see that b is the m—intercept for the tangent line at (a + c) / 2 and a is the sic—intercept for the tangent line at (b + c)/2. #1: (x 7 al-(x — bl-lx ~ C) Author the function y: D a + b #2: TANGENT (x — a)~lx ~ b)«(x — c). x, ~—4———~r Tangent(#l. x (a+bl/2l 2 2 2 (a — 2~a-b + b ) (c — x) #3: OlOs Simp(#2) 4 ...
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